Question: Simplify and expand the following expression: $ \dfrac{5}{5z + 5}- \dfrac{4}{3z - 6}- \dfrac{2z}{z^2 - z - 2} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{5}{5z + 5} = \dfrac{5}{5(z + 1)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{4}{3z - 6} = \dfrac{4}{3(z - 2)}$ We can factor the quadratic in the third term: $ \dfrac{2z}{z^2 - z - 2} = \dfrac{2z}{(z + 1)(z - 2)}$ Now we have: $ \dfrac{5}{5(z + 1)}- \dfrac{4}{3(z - 2)}- \dfrac{2z}{(z + 1)(z - 2)} $ The least common multiple of the denominators is: $ 15(z + 1)(z - 2)$ In order to get the first term over $15(z + 1)(z - 2)$ , multiply by $\dfrac{3(z - 2)}{3(z - 2)}$ $ \dfrac{5}{5(z + 1)} \times \dfrac{3(z - 2)}{3(z - 2)} = \dfrac{15(z - 2)}{15(z + 1)(z - 2)} $ In order to get the second term over $15(z + 1)(z - 2)$ , multiply by $\dfrac{5(z + 1)}{5(z + 1)}$ $ \dfrac{4}{3(z - 2)} \times \dfrac{5(z + 1)}{5(z + 1)} = \dfrac{20(z + 1)}{15(z + 1)(z - 2)} $ In order to get the third term over $15(z + 1)(z - 2)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{2z}{(z + 1)(z - 2)} \times \dfrac{15}{15} = \dfrac{30z}{15(z + 1)(z - 2)} $ Now we have: $ \dfrac{15(z - 2)}{15(z + 1)(z - 2)} - \dfrac{20(z + 1)}{15(z + 1)(z - 2)} - \dfrac{30z}{15(z + 1)(z - 2)} $ $ = \dfrac{ 15(z - 2) - 20(z + 1) - 30z} {15(z + 1)(z - 2)} $ Expand: $ = \dfrac{15z - 30 - 20z - 20 - 30z}{15z^2 - 15z - 30} $ $ = \dfrac{-35z - 50}{15z^2 - 15z - 30}$ Simplify: $ = \dfrac{-7z - 10}{3z^2 - 3z - 6}$